java - 如何计算两个 java.sql.Time 值之间的差异？

Question

我有如下表格，

id | Lunch_Out | After_Lunch_In

01   |  01:15:00   |   02:00:01

我正在尝试使用以下代码查找时差，

while(rst.next())
            {
                PrintWriter obj1 = response.getWriter();
                obj1.println("while entered");

                Time a =rst.getTime("Lunch_Out");
                Time b =rst.getTime("After_Lunch_In");

                PrintWriter objt1 = response.getWriter();
                objt1.println("LougOut Time is :"+b);
                PrintWriter objt2 = response.getWriter();
                objt2.println("LogIn Time is :"+a);

//long c = b.getTime() - a.getTime() 

                long c = b.getTime() - a.getTime() / (24 * 60 * 60 * 1000);

                //PrintWriter objtr = response.getWriter();
                //objtr.println("different  is :"+c);

                Time diff = new Time(c);

                PrintWriter objt = response.getWriter();
                objt.println("different  is :"+diff);

            }

输出是:

while entered

LougOut Time is :02:00:02

LogIn Time is :01:15:01

different  is :02:00:02

但预期输出是:00:45:01。我在哪里做错了？

Answer 1

您正在将 a 和 b 都转换为毫秒(从纪元开始)，然后求出差异。由于您的后续行涉及调用 Time constructor ，因此您不需要像您所做的那样通过除以 24 * 60 * 60 * 1000 来将时间扩展到几天。仅将c 值作为毫秒差值，然后将其转换为下一行中的Time 对象diff。

更新后的代码应如下所示。

long c = b.getTime() - a.getTime();
Time diff = new Time(c);

但是，这将为您提供一个具有正确小时数、分钟数和秒数但日期为 1970 年 1 月 1 日的 Time 对象。由于您已将 MySQL 标记添加到您的问题中，我建议您考虑 Paul '在他的评论中提出了建议。

I suggest you let MySQL do the math for you as part of your query, something like SELECT Lunch_Out, After_Lunch_In, subtime(After_Lunch_In, Lunch_Out) as Lunch_Duration FROM .... See the MySQL reference manual for details.